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arlik [135]
3 years ago
14

Rita offers to pay for 8 of her friends to play lazer tag, but 3 of them dont want to pay

Mathematics
2 answers:
Mrac [35]3 years ago
6 0

Answer:

only 5 of her friends will play

Step-by-step explanation:

8-3=5

boyakko [2]3 years ago
5 0

Answer:

The answer 5

Step-by-step explanation:

8 friends minus the three that dont want to play

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A botanist transplanted a plant that
KengaRu [80]

Linear equation of plant height is h = (1/2)w + 3

Given:

Height of experimental plant = 3cm

Average growth = 0.5 cm per week

Find:

Linear equation of plant height

Computation:

Given;

h = Plant height

w = number of weeks

So,

Plant height = Height of experimental plant + (Average growth)(number of weeks)

h = 3 + (0.5)(w)

h = (1/2)w + 3

Option "J" is the correct answer to the following question.

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2 years ago
If 4 is added to 8 to equal 12, how would the operation be undone​
yarga [219]

Answer:

4+8=12

Step-by-step explanation:

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2 years ago
Jake lives in San Mateo. He bought an item for $787.31 and when the clerk calculated the sales tax it came out to exactly
Tems11 [23]

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6 0
2 years ago
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
Find the solution of this system of equations.
Elenna [48]
<span>-4x + y = -25        y = 4x - 25

-6x - 6y = 0            </span>
 
-6x - 6(4x - 25) = 0
-6x - 24x + 150 = 0
-30x + 150 = 0
-30x = -150
x = 5

y = 4(5) - 25
y = 20 - 25
y = -5

(5, -5)
                       
8 0
3 years ago
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